Question

@ganeshie8 @hartnn

Answer 1

If alpha and beta are the roots of the equation ax^2 + bx + c=0 find the equation whose roots are \[\huge \frac{ 1+\alpha }{ 1-\alpha } , \frac{ 1+\beta }{ 1-\beta } \]

Answer 2

sum of roots = alpha + beta =....
product of roots = alpha*beta =...

in terms of a,b,c

Answer 3

-b/a
c/a

Answer 4

ok so i add the fractions and multiply the fractions

Answer 5

correct

Answer 6

x^2 - (Sum of roots)x + (product of roots )

Answer 7

I am not getting it after i multiplied wait i will show you what i did mean while u can answer other questions it would take some time to type

Answer 8

I am getting this when i add these
\[\huge \frac{ 2(1-\frac{ c }{ a }) }{ \frac{ c }{ a }-\alpha -\beta +1 }\]

And after multiplying :-
\[\huge \frac{ 1 - \frac{ b }{ a }+ \frac{ c }{ a } }{ \frac{ c }{ a } -\alpha -\beta +1 }\]

Answer 9

plug in alpha + beta as -b/a

Answer 10

those are correct
just plug in
-alpha - beta
as b/a

Answer 11

MY net is down so i replied a bit late

Answer 12

thats okay :)

Answer 13

\[x ^{2} - \left( \huge \frac{ 2(1-\frac{ c }{ a }) }{ \frac{ c }{ a } + \frac{ b }{ a } +1 } \right)x + \left( \huge \frac{ 1 - \frac{ b }{ a }+ \frac{ c }{ a } }{ \frac{ c }{ a}+\frac{ b }{ a } +1 }\right)=0\]

Answer 14

Phew!

Answer 15

yesss

simplify it

first multiply numerator and denominator of denominators by 'a'
then
multiply throughout by a+b+c

Answer 16

see if you get
\(\large (a+b+c)x^2-2(a-c)x+(a-b+c)=0\)
in the end.